Evaluate the following Integrals:

Let


As we have the trigonometric identity



to evaluate this integral we use x = acos 2θ


dx = –2a sin(2θ) dθ (Differentiating both sides)


When x = –a, acos 2θ = –a cos 2θ = –1



When x = a, acos 2θ = a cos 2θ = 1


2θ = 0 θ = 0


So, the new limits are and 0.


Also,





Substituting this in the original integral,






But, we have 2 sin2θ = 1 – cos 2θ









55